Let a compact Lie group $G$ act on a manifold $M$. Fix $X \in \mathfrak g$ and we write $X_M$ for the infinitesimal action of $X$. Assume that $p \in M$ is a zero point of $X_M$.
Define a linear map $L_p : T_pM \to T_pM$ by $L_p(v) = \mathcal L(X_M)v$. Here $v$ on the right hand side is an extension of the tangent vector $v$ to a vector field. Since $X_M(p)=0$, $L_p$ is well-defined.
I want to show that $L_p$ is invertible if $p$ is an isolated zero point. Here is a proof (Lemma 45).
Proof. Pick a $G$-invariant Riemannian metric and suppose that $v \in T_pM-0$ is annihilated by $L_p$. Taking the exponential map with respect to the metric, we see that all the points on the geodesic $\exp_p (tv)$, $t \in \mathbb R$, would be fixed by $\exp(sX) \in G$, $s \in \mathbb R$, contradicting to $p$ being isolated.
(I changed the proof slightly due to notation.)
Could you tell me why $\exp_p(tv)$ is fixed by $\exp(sX)$?
$\exp(sX)$ preserves the metric, so it sends geodesics to geodesics. As a geodesic is uniquely specified by its initial position and velocity, $\exp(sX)$ sends the geodesic $\exp_p(tv)$ to itself.